The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is -

Let the 3 consecutive number is
x, x + 1, x + 2
According to question,
$$ \Rightarrow {x^2} + {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2}$$     = 365
$$ \Rightarrow {x^2} + {x^2} + 1 + 2x + {x^2} + 4 + 4x$$       = 365
$$\eqalign{ & \Rightarrow 3{x^2} + 6x = 360 \cr & \Rightarrow {x^2} + 2x - 120 = 0 \cr & \Rightarrow \left( {x - 10} \right)\left( {x + 12} \right) = 0 \cr & \Rightarrow x = 10 \cr & {\text{Sum of numbers :}} \cr & = 10 + 11 + 12 \cr & = 33 \cr} $$